= Solution
It helps to separate the <tangent> and the squaring map. Put $t=\tan(z/2)$, with $z=x+iy$. The identity
$$
t=\frac{\sin x+i\sinh y}{\cos x+\cosh y},\qquad |t|^2=\frac{\cosh y-\cos x}{\cosh y+\cos x}
$$
shows that $0<x<\pi/2$ gives $\operatorname{Re}t>0$ and $|t|<1$. To see that every point of this right half-disc occurs, set
$$
R=\frac{1+it}{1-it},\qquad \operatorname{Re}R=\frac{1-|t|^2}{|1-it|^2}>0,\qquad \operatorname{Im}R=\frac{2\operatorname{Re}t}{|1-it|^2}>0.
$$
Then $z=-i\operatorname{Log}R$ has $0<\operatorname{Re}z<\pi/2$ and satisfies $\tan(z/2)=t$. The <principal complex logarithm> is unambiguous because $R$ is in the first quadrant.
Squaring maps this right half-disc bijectively onto the <unit disc> with the interval $(-1,0]$ removed: the argument of $t$ lies between $-\pi/2$ and $\pi/2$, and doubles into $(-\pi,\pi)$. Thus \b[the desired image is]
$$
\boxed{\{w:|w|<1\}\setminus(-1,0].}
$$
Neither derivative vanishes in the domain, so the composition is also a <conformal map> onto that slit disc.
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